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dangina [55]
3 years ago
13

A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3

kg moving in the negative X-Direction with a speed of 1 m/s . During collision a loud sound is heard and they both start moving together . the sound energy cannot be greater than :
A) 12 J
B) 14 J
C) 15 J
D) 17.5 J 
Physics
1 answer:
Inga [223]3 years ago
8 0
m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.

The momentul of the system preserves:

m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with E.

Now, we write the energy conservation law:

\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E

From the above equation, you find E,  and then conclude that the sound energy can certainly not be greater than this.
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A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
What is the volume of a rock with a density of 3.00 g/cm3 and a mass of 600g?
Mademuasel [1]
The equation of D = m/V

Where D = density
m = mass
and V = volume

We are solving for V, so with the manipulation of variables we multiply V on both sides giving us 
V(D) = m 
now we divide D on both sides giving us
V = m/D 

We know our mass which is 600g and our density is 3.00 g/cm^3
so
V = 600g/3.00g/cm^3 = 200cm^3  or 200mL

a cubic centimeter (cm^3) is one of the units for volume. It's exactly like mL. 1 cm^3 = 1 mL
 
If you wish to change it to L, you'd have to convert. 
5 0
3 years ago
Describe the initial horizontal and vertical velocity of a horizontally launched projectile on Earth, as well as what happens to
Jobisdone [24]

Answer:

Explained below

Explanation:

To explain this, let's consider a tennis ball being launched from the top of a very high building.

Now, if the tennis ball is launched horizontally without any upward angle but with an initial velocity of 10 m/s. In this motion, If there is no gravity, the tennis ball would continue in motion at that same speed of 10 m/s in the horizontal direction. However, in reality, gravity causes the tennis ball to accelerate downwards at a rate of 9.8 m/s for every second. This implies that the vertical velocity component is changing at the rate of 9.8 m/s every second.

Thus, after 1 second, horizontal velocity component will remain 10 m/s and vertical component will be 9.8 m/s × 1 = 9.8 m/s downwards.

Also, after 2 seconds, the vertical velocity component will remain 10 m/s, however the vertical component will now be 9.8 × 2 = 19.6 m/s downwards.

Same procedure is repeated as t increases by 1 second.

5 0
3 years ago
PLS SOMEONE HELP QUICK
SVETLANKA909090 [29]

Answer:

the rates of rock formation are similar. i could be wrong tho.....

Explanation:

6 0
2 years ago
Read 2 more answers
A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats
Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

\lambda = \frac{c}{f}

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m            

Hence, the distance is:

d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

 

I hope it helps you!

7 0
3 years ago
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